Evaluate $~~\int^\pi_{3\pi/2} x\cos x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $\pi$ (Choice C) C $\dfrac{3\pi}2-\pi$ (Choice D) D $\dfrac{3\pi}2-1$ (Choice E) E $\dfrac{3\pi}2+1$
We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = x~$ and $~dv=\cos x \,dx\,$. Then $~du = dx~$ and $~v = \sin x\,$. Integration by parts gives $ \int^\pi_{3\pi/2} x\cos x\,dx = x\sin x-\int^\pi_{3\pi/2}\sin x\,dx$ $ ~=x\sin x+\cos x\Bigg]^\pi_{3\pi/2}$ $ ~=\pi\sin\pi+\cos\pi-\Big(\frac{3\pi}2\sin\frac{3\pi}2+\cos\frac{3\pi}2\Big)$ $ ~=0-1-\Big(-\dfrac{3\pi}{2}+0\Big) =\dfrac{3\pi}2-1$